Problem: $ \lim_{x\to 2}\sqrt{6x+13}=$
Explanation: $\sqrt{6x+13}$ defines a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$ -values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim_{x\to c}q(x)=q(c)$ [What happens at the endpoint?] The input $x=2$ is within the domain of $\sqrt{6x+13}$. Therefore, in order to find $ \lim_{x\to 2}\sqrt{6x+13}$, we can simply evaluate $\sqrt{6x+13}$ at $x=2$. $\begin{aligned} &\phantom{=}\sqrt{6x+13} \\\\ &=\sqrt{6(2)+13} \gray{\text{Substitute }x=2} \\\\ &=\sqrt{25} \\\\ &=5 \end{aligned}$ In conclusion, $ \lim_{x\to 2}\sqrt{6x+13}=5$.